**Background**

Suppose that we are using a simplified spherical model of the Earth’s surface with latitude $u in (-frac {pi} 2, frac {pi} 2)$ and longitude $v in (-pi, pi)$; then (if the radius is taken to be $1$), the surface area element is given by $mathrm{d}A = cos u mathrm{d}u mathrm{d}v$. Restricting attention to the hemisphere, $H$, where $u, v in (-frac {pi} 2, frac {pi} 2)$, a simple map projection from $H$ can be obtained by just taking the $x$ and $y$ coordinates via $x = cos u sin v$ and $y = sin u$, which is a smooth one-to-one transformation on $H$. Now, picking a point with coordinates $(U, V)$ on $H$ uniformly according to surface area, the joint density of $U$ and $V$ is $$f_{U, V}(u, v) = frac 1 {2pi} cos u, quad lvert u rvert, lvert v rvert < frac {pi} 2.$$

**Question**

$(a)quad$ Find $mathbb{E}(lvert sin U rvert mid V = 0)$.

$(b)quad$ Find $mathbb{E}(lvert Y rvert mid X = 0)$.

$(c)quad$ Observe that $lvert Y rvert = lvert sin U rvert$ and the event ${X = 0}$ is exactly the same as the event ${V = 0}$. How is it possible that $mathbb{E}(lvert Y rvert mid X = 0) neq mathbb{E}(lvert sin U rvert mid V = 0)$?

**My working**

I have omitted intermediate steps and only shown the essential parts.

$(a)$

$$begin{aligned}

because f_{U mid V = v}(u) & = frac 1 2 cos u,quad lvert u rvert, lvert v rvert < frac pi 2

\(5 mm) therefore mathbb{E}(lvert sin U rvert mid V = 0) & = int^{infty}_{-infty} lvert sin u rvert left(frac 1 2 cos uright) mathrm{d}u

\(5 mm) & = int^{frac pi 2}_0 sin u cos u mathrm{d}u

\(5 mm) & = frac 1 2

end{aligned}$$

$(b)$

$$begin{aligned}

\(5 mm) because f_{X, Y}(x, y) & = frac 1 {2 pi sqrt{1 – y^2 – x^2}}, quad x^2 + y^2 < 1

\(5 mm) therefore f_{Y mid X = x}(y) & = frac {frac 1 {2 pi sqrt{1 – y^2 – x^2}}} {int^{sqrt{1 – x^2}}_{-sqrt{1 – x^2}} frac 1 {2 pi sqrt{1 – y^2 – x^2}} mathrm{d}y}

\(5 mm) & = frac 1 {pi sqrt{1 – y^2 – x^2}}, quad x^2 + y^2 < 1

\(5 mm) implies mathbb{E}(lvert Y rvert mid X = 0) & = int^{infty}_{-infty} frac {lvert y rvert} {pi sqrt{1 – y^2}} mathrm{d}y

\(5 mm) & = frac 2 pi int^1_0 frac y {pi sqrt{1 – y^2}} mathrm{d}y

\(5 mm) & = frac 2 pi

end{aligned}$$

I think my answers to $(a)$ and $(b)$ are correct, but I am not sure how to approach $(c)$. Any intuitive explanations will be greatly appreciated! Also, if my answers to $(a)$ and/or $(b)$ are wrong, please do point out my mistakes ðŸ™‚